Before answering the questions, lets first find out what adding the heat does. Initially the ice is at -20 C and must be warmed to 0C before it will melt. This will take an amount of heat: q = (mass)C_SDT = (100g)( 1.95 J/( K g))(20K) = 3900 J. So we used up 3.90 kJ warming the ice up to the temperature of fusion (the melting point). That means we have 16.10 kJ left to use in melting the ice. How much ice can we melt with 16.10 kJ? Well, for the phase change q = nDH_fus. n = q/ DH_fus = (16.10 kJ)/(6.008 kJ/mole) = 2.680 moles which is just (2.680 moles)(18 g/mole) = 48.24 grams of ice melted into water. Since the density of liquid water is 1.00 grams/ml, this means that after adding the heat we have 48.24 mls of liquid water and 51.76 grams of ice. So the answers are:

a) 0 C because we have both ice and water at equilibrium in the end.

b) 48.2 mls of water and 51.8 grams of ice, as described above.

c) For the heating of the ice to 0 C, the entropy change is

C_S(mass)ln(T_f/T_I) = (1.95 J/(K g))(100 g)ln(273.15/253.15) = 14.83 J/K

For the melting of the ice, the entropy change is

n DH_fus/T_fus = (48.24/18)(6.008 kJ/mole)/(273.15 K) = 58.95 J/K

So the entropy change for the whole process (the entropy change of the system) is 73.8 J/K